Two hydrogen and two oxygen multiply for two
Answer:
waste gas he should use algae
Answer:
3). 1.30 × 10^(24) molecules
Explanation:
From avogadro's law which state that equal volume of all gases at the same temperature and pressure contain the same number of molecules.
We can relate it to this question as;
V₁/n₁ = V₂/n₂
Where;
V₁ is initial volume
n₁ is initial number of molecules
V₂ is final volume
n₂ is final number of molecules
Thus at STP, we have V₁ = V₂ and as such Plugging in the relevant values gives;
5/(1.30 x 10^(24)) = 5/n₂
n₂ = 1.30 x 10^(24) molecules
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
A straight line on a distance va time graph represents constant speed
